Ratio test for series

[woopydalan]

Homework Statement:

Use the ratio test for the series $\sum_{n=1}^{\infty} b_{n}$ where $b_{1} = 5$ and $b_{n}= \frac {(-1)^{n}nb_{n-1}}{3n+1}$ for $n \geq 2$

Relevant Equations:

ratio test: $L = \lim_{n \to \infty} \lvert \frac {a_{n+1}}{a_{n}} \rvert$ will converge if $L < 1$ and will diverge if $L > 1$

So I am having some difficulty expressing this series explicitly. I just tried finding some terms

$b_{0} = 5$

I am assuming I am allowed to use that for $b_{1}$ for the series, even if the series begins at $n=1$? With that assumption, I have

$b_{1} = -\frac {5}{4}$
$b_{2} = - \frac{5}{14}$
$b_{3} = \frac {3}{28}$
$b_{4} = \frac {3}{91}$

I am not seeing a pattern. I also alternatively tried using the ratio test not in explicit terms

$\lim_{n \to \infty} \lvert \frac {(-1)^{n+1}(n+1)b_{n}}{3(n+1)+1} \cdot \frac {3n+1}{(-1)^{n}nb_{n-1}} \rvert$ and after substituting I ended up with $\lvert \frac{b_{n}}{b_{n-1}} \rvert$ which to me is inconclusive.

 

[vela]

Homework Statement:: Use the ratio test for the series $\sum_{n=1}^{\infty} b_{n}$ where $b_{1} = 5$ and $b_{n}= \frac {(-1)^{n}nb_{n-1}}{3n+1}$ for $n \geq 2$
Relevant Equations:: ratio test: $L = \lim_{n \to \infty} \lvert \frac {a_{n+1}}{a_{n}} \rvert$ will converge if $L < 1$ and will diverge if $L > 1$

So I am having some difficulty expressing this series explicitly. I just tried finding some terms

You're making this way too complicated.
$$b_{n}= \frac {(-1)^{n}nb_{n-1}}{3n+1} \quad \Rightarrow \quad \frac{b_{n}}{b_{n-1}}= \frac {(-1)^{n}n}{3n+1}$$

 

[woopydalan]

Oh wow...let me try that

 

[woopydalan]

Ok, so now I got $\lim_{n \to \infty} \frac {n}{3n+1}$ which would be indeterminate??

 

[vela]

Do you mean indeterminate in the sense that the limit is $\infty/\infty$ or do you mean the convergence of the series can't be determined?

 

[woopydalan]

I think in both senses

 

[pasmith]

Hint:
$$\lim_{n \to \infty} \frac{an+b}{cn+d} = \lim_{n \to \infty} \frac{a + \frac{b}{n}}{c + \frac{d}{n}} = \frac{a}{c}$$ for $c \neq 0$.

 

[woopydalan]

Alright, so 1/3, so L < 1, so converges